Difference between revisions of "2001 AMC 8 Problems/Problem 23"

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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20</math>
 
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20</math>
  
==Solution==
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==Solution 1==
There are <math> 6 </math> points in the figure, and <math> 3 </math> of them are needed to form a triangle, so there are <math> {6\choose{3}} =20 </math> possible triples of <math> 3 </math> of the <math> 6 </math> points. However, some of these created congruent triangles, and some don't even make triangles at all.
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There are <math> 6 </math> points in the figure, and <math> 3 </math> of them are needed to form a triangle, so there are <math> {6\choose{3}} =20 </math> possible triplets of the <math> 6 </math> points. However, some of these created congruent triangles, and some don't even make triangles at all.
  
 
'''Case 1: Triangles congruent to <math> \triangle RST </math>''' There is obviously only <math> 1 </math> of these: <math> \triangle RST </math> itself.
 
'''Case 1: Triangles congruent to <math> \triangle RST </math>''' There is obviously only <math> 1 </math> of these: <math> \triangle RST </math> itself.
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However, if we add these up, we accounted for only <math> 1+4+6+6=17 </math> of the <math> 20 </math> possible triplets. We see that the remaining triplets don't even form triangles; they are <math> SYR, RXT, </math> and <math> TZS </math>. Adding these <math> 3 </math> into the total yields for all of the possible triplets, so we see that there are only <math> 4 </math> possible non-congruent, non-degenerate triangles, <math> \boxed{\text{D}} </math>
 
However, if we add these up, we accounted for only <math> 1+4+6+6=17 </math> of the <math> 20 </math> possible triplets. We see that the remaining triplets don't even form triangles; they are <math> SYR, RXT, </math> and <math> TZS </math>. Adding these <math> 3 </math> into the total yields for all of the possible triplets, so we see that there are only <math> 4 </math> possible non-congruent, non-degenerate triangles, <math> \boxed{\text{D}} </math>
== easy solution==
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Note that it has to be less than 6C3 because some groups of three points make lines.  
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==Solution 2==
One possible triangle is the one connecting all the midpoints, another is the original equilateral triangle, a third one is a right triangle and the fourth one is an obtuse triangle by connecting two midpoints and a vertex that is not the vertex between them. Because there are no more answer choices from 5-19, our answer is 4.
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We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is <math>3+1 = \boxed{4}</math>, which is <math>\boxed{\text{D}}</math>
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-FIREDRAGONMATH16
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==Solution 3 (Multiple-choice abuse)==
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Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: <math>\triangle RYX</math>, <math>\triangle RYT</math>, <math>\triangle RYZ</math>, <math>\triangle RST</math>. So the answer is <math>\boxed{\text{(D) } 4}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=22|num-a=24}}
 
{{AMC8 box|year=2001|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Combinatorics Problems]]

Revision as of 16:09, 13 July 2021

Problem

Points $R$, $S$ and $T$ are vertices of an equilateral triangle, and points $X$, $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?

[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution 1

There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triplets of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.

Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself.

Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$.

Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$.

Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$.

However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$. Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{\text{D}}$

Solution 2

We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is $3+1 = \boxed{4}$, which is $\boxed{\text{D}}$

-FIREDRAGONMATH16

Solution 3 (Multiple-choice abuse)

Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$, $\triangle RYT$, $\triangle RYZ$, $\triangle RST$. So the answer is $\boxed{\text{(D) } 4}$

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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